2.1. INTRODUCTORY EXAMPLES: PENDULUM PROBLEMS 23

represents small deviations around the vertical position. For this situation, we
can use the approximationscosθ=− 1 ,sinθ=−φ,and®θ=0. By substituting
these approximations into Equations 2.15 and 2.16, we obtain

(M+m)®x+bx ̇−mLφ®=u (2.17)

and

(mL^2 +I)®φ−mgLφ=mL®x (2.18)

Equations 2.17 and 2.18 are the linearized set of equations we will use to design
the PID controller.
Wefirst derive the transfer function for the inverted pendulum. To do this,
we take the Laplace transform of Equations 2.17 and 2.18 with zero initial
conditions which yields

(M+m)s^2 X(s)+bsX(s)−mLs^2 Φ(s)=U(s) (2.19)

and

(mL^2 +I)s^2 Φ(s)−mgLΦ(s)=mLs^2 X(s) (2.20)

Since we are interested in the deviationΦ(s)in the pendulum from the
vertical position, as a function of the stateX(s), we solve Equation 2.20 for
X(s)to obtain

X(s)=

"°

mL^2 +I

¢

mL

−

g

s^2

Φ(s) (2.21)

Substituting Equation 2.21 into 2.19, we obtain the relationship betweenΦ(s)
and the stateX(s)as

(M+m)

∑

(mL^2 +I)

mL +

g

s

∏

s^2 Φ(s)+b

∑

(mL^2 +I)

mL +

g

s

∏

sΦ(s)−mLs^2 Φ(s)
=U(s)
(2.22)
Rearranging Equation 2.22, we obtain the transfer function

Φ(s)

U(s)

=

mL

r s

2

s^4 +b(mL

(^2) +I)
r s
(^3) −(M+m)mgL
r s
(^2) −bmgL
r s

mL
rs
s^3 +b(mL
(^2) +I)
r s
(^2) −(M+m)mgL
r s−
bmgL
r
where
r=
h
(M+m)

°

mL^2 +I

¢

−(mL)^2

i

Using the method outlined earlier, the linearized equations may be expressed